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If $a,b,c \in R$, then prove that:

$$\frac{bc}{b+c}+\frac{ac}{a+c}+\frac{ab}{a+b} \leq \frac{a+b+c}{2}$$

I can't see any known inequality working here like $H.M.-A.M.$. Could this be solved using basic inequalities?

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    $\begingroup$ I suspect that $a,b,c$ must be greater than zero. Cause the inequality fails for $a=-1$, $b=2$, $c=-3$. $\endgroup$ – Ángel Mario Gallegos Apr 16 '16 at 19:21
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For $a,b,c>0$ we get
$(a-b)^2=(a+b)^2-4ab\geq 0\implies \frac{ab}{a+b}\leq \frac{a+b}{4}$, the equality occurs only when $a=b$.

In similar way we have
$\frac{bc}{b+c}\leq \frac{b+c}{4}$ and $\frac{ca}{c+a}\leq \frac{c+a}{4}$
Adding these we get $$\frac{bc}{b+c}+\frac{ac}{a+c}+\frac{ab}{a+b} \leq \frac{2(a+b+c)}{4}=\frac{a+b+c}{2}$$

The equality occurs only when $a=b=c$

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Actually HM-AM works here: for any $x,y>0$ we have $$\frac{2xy}{x+y} = \frac{2}{\frac 1x + \frac 1y} \le \frac{x+y}2.$$

Use the inequality above for $(x,y) \in \{(b,c), (c,a), (a,b)\}$, sum them up, and divide by 2.

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it holds for $$a,b,c>0$$ and is equivalent to $$b(a-c)(a^2-c^2)+c(a-b)(a^2-b^2)+a(b-c)(b^2-c^2)\geq 0$$ which is true.

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  • $\begingroup$ because it can be writtenb(a-c)(a^2-c^2)+c(a-b)(a^2-b^2)+a(b-c)(b^2-c^2)\geq 0 $\endgroup$ – Jean Marie Apr 16 '16 at 19:27
  • $\begingroup$ yes you have right $\endgroup$ – Dr. Sonnhard Graubner Apr 16 '16 at 19:28
  • $\begingroup$ ....because it can be written $b(a-c)^2(a+c)+c(a-b)^2(a+b)+a(b-c)^2(b+c)$ which is $\geq 0$ with equality to $0$ iff $a=b=c$ (under the hypothese of positivity for $a,b$ and $c$. $\endgroup$ – Jean Marie Apr 16 '16 at 19:30
  • $\begingroup$ yes this can also be written in the form above $\endgroup$ – Dr. Sonnhard Graubner Apr 16 '16 at 19:32

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